A block with mass m =6.3 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.3 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.26 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5 m/s. The block oscillates on the spring without friction.

1) What is the spring constant of the spring?

2) What is the oscillation frequency?

3) After t = 0.38 s what is the speed of the block?

4) What is the magnitude of the maximum acceleration of the block?

5) At t = 0.38 s what is the magnitude of the net force on the block?

6) Where is the potential of the system the greatest?

Explanation / Answer

here

at equilibrium

mg=kx

1) k=6.3*9.81/0.26=237.7 N/m

and w=sqrt(k/m)=6.142 rad/s

2) f=w/(2pi)=0.977 Hz

let equation of motion be

x(t)=Asin(wt)

initial vel =5

so x'=Awcos(wt)

so Aw=5

so A=0.814 m

3) x'(0.38)=0.814*6.142*cos(6.142*0.38)=-3.455 m/s

4) max acc=x''=Aw^2=0.814*6.142^2=30.7 m/s^2

5) a(0.38)=x''(0.38)=Aw^2sin(wt)=30.7*sin(6.142*0.38)=22.185 m/s^2

6) the potential of the system is greatest when the velocity is zero ..at the max amplitude point

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.